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podryga [215]
3 years ago
10

Could someone help me with this problem?

Mathematics
1 answer:
Alisiya [41]3 years ago
3 0
The answer is root 65
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Foster is centering a photo that is 1/1 2 inches wide on a scrapbook page that is 10 inches wide. How far from each side of the
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The answer would be 4 1/4 (4.25)
10-1.5=8.5/2=4.25

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Choose the linear inequality that describes the graph. The gray area represents the shaded region
Vesnalui [34]
(0,3)(1,-2)
slope = (-2-3) / (1 - 0) = -5

y int is (0,3)

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ur inequality is : y > -5x + 3


6 0
3 years ago
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A traveling soccer team allows only the top 3% of athletes who try out to be part of the team. If the team tryout scorecard has
skad [1K]

Answer:

169

Step-by-step explanation:

You need to use a Z-table for this.

There are different tables, in this table i searched for the probability of 97% because that equivalent to the top 3%.

for 0.97, Z = 1.88

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Z = (x-mean)/(st. dev.)

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1.88 = (x-150)/10 \\x = 168.8 = 169

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3 years ago
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Answer:

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3 years ago
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HI PLEASE HELP ME WITH MY CALCULUS 1 HW? I AM REALLY STUCK. I need help with parts d,e,g.
asambeis [7]

(d) The particle moves in the positive direction when its velocity has a positive sign. You know the particle is at rest when t=0 and t=3, and because the velocity function is continuous, you need only check the sign of v(t) for values on the intervals (0, 3) and (3, 6).

We have, for instance v(1)\approx-0.91 and v(4)\approx0.91>0, which means the particle is moving the positive direction for 3, or the interval (3, 6).

(e) The total distance traveled is obtained by integrating the absolute value of the velocity function over the given interval:

\displaystyle\int_0^6|v(t)|\,\mathrm dt=\int_0^3-v(t)\,\mathrm dt+\int_3^6v(t)\,\mathrm dt

which follows from the definition of absolute value. In particular, if x is negative, then |x|=-x.

The total distance traveled is then 4 ft.

(g) Acceleration is the rate of change of velocity, so a(t) is the derivative of v(t):

a(t)=v'(t)=-\dfrac{\pi^2}9\cos\left(\dfrac{\pi t}3\right)

Compute the acceleration at t=4 seconds:

a(t)=\dfrac{\pi^2}{18}\dfrac{\rm ft}{\mathrm s^2}

(In case you need to know, for part (i), the particle is speeding up when the acceleration is positive. So this is done the same way as part (d).)

6 0
4 years ago
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